Question

Determine the zeros of f(x) = x^3 – 3x^2 – 16x + 48

Answer 1

you just have to factor the equation to easily see the zeros
factoring it gives \[(x ^{2}-16)*(x-3)\]= \[(x +4)(x -4)(x-3)\]
When that equals zero, x has to be 4, -4, or 3 to cause the equation to be true.

Answer 2

Ah. Thanks!